FIELD OF p-ADIC NUMBERS
J.M. CASAS, B.A.OMIROV, U.A. ROZIKOV
Abstract. We establish the solvability criteria for the equationxq =ain the field of p-adic numbers, for any q in two cases: (i) q is not divisible by p; (ii) q = p.
Using these criteria we show that anyp-adic number can be represented in finitely many different forms and we describe the algorithms to obtain the corresponding representations. Moreover it is showed that solvability problem ofxq =afor anyq can be reduced to the cases (i) and (ii).
AMS classifications (2010): 11S05
Keywords: p-adic number; solvability of an equation; congruence.
1. Introduction
Thep-adic number system for any prime numberpextends the ordinary arithmetic of the rational numbers in a way different from the extension of the rational number system to the real and complex number systems. This extension is achieved by an alternative interpretation of the concept of absolute value.
First described by Kurt Hensel in 1897, the p-adic numbers were motivated pri- marily by an attempt to bring the ideas and techniques of power series methods into number theory. Their influence now extends far beyond this. For example, the field of p-adic analysis essentially provides an alternative form of calculus.
More formally, for a given primep, the fieldQpofp-adic numbers is a completion of the rational numbers. On the fieldQp is also given a topology derived from a metric, which is itself derived from an alternative valuation on the rational numbers. This metric space is complete in the sense that every Cauchy sequence converges to a point inQp. This is what allows the development of calculus onQp and it is the interaction of this analytic and algebraic structure which gives the p-adic number systems their power and utility.
For about a century after the discovery of p-adic numbers, they were mainly con- sidered as objects of pure mathematics. However, numerous applications of these numbers to theoretical physics have been proposed in papers [1, 19, 5, 10], to quan- tum mechanics [8], top-adic - valued physical observables [8] and many others [7, 18].
As in real case, to solve a problem inp-adic case there arises an equation which must
1
be solved in the field ofp-adic numbers (see for example [11, 12, 13, 18]). For classi- fication problems of varieties of algebras over a field Qp of p-adic numbers one has to solve an equation of the form: x2 = a. It is well known the criteria of solvability of this equation. In fact, in the classification of Leibniz algebras of dimensions less than 4 over a field Qp (see [3], [9]) it is enough to solve the equation x2 = a. However, in the classifications tasks of larger dimensions one has to solve an equation of the form xq = a, q ≥ 2, in Qp. In this paper we present a criteria for solvability of the equationxq =ainQp (wherepis a fixed prime number) for arbitrary q in two cases:
(q, p) = 1 and q=p. Moreover, in the cases of existing the solutions of the equation we present the algorithm of finding the solutions. Also we show that any equation xq =a inQp can be reduced to both cases. Note that in [14] the same criterion has been proved by a different method.
2. Preliminaries
2.1. Solvability of congruences. The sources of the information in this subsection are [2, 4, 15]. If n is a positive integer, the integers between 1 and n−1 which are coprime ton(or equivalently, the congruence classes coprime ton) form a group with multiplication modulo nas the operation; it is denoted byZ×n and is called the group of units modulon or the group of primitive classes modulon. Multiplicative group of integers modulo n is cyclic if and only if n is equal to 1,2,4, pk or 2pk, where pk is a power of an odd prime number. A generator of this cyclic group is calleda primitive root modulo n or a primitive element of Z×n.
For any integers a, b and n, we say that a is congruent to b modulo n (notated a ≡ b mod n) if n | (b −a). The order of Z×n is given by Euler’s totient function ϕ(n). Euler’s theorem says that aϕ(n) ≡ 1 mod n for every a coprime to n; the smallest k for which ak ≡1 mod n is called the multiplicative order of a modulo n.
In other words, for a to be a primitive root modulo n, ϕ(n) has to be the smallest power of a which is congruent to 1 modn.
Lemma 2.2. Suppose that m ∈ N has a primitive root r. If a is a positive integer with (a, m) = 1, then there is a unique integer x with 1≤x≤ϕ(m) such that
rx ≡a modm.
Definition 2.3. If m ∈ N has a primitive root r and α is a positive integer with (α, m) = 1, then the unique integer x, 1 ≤ x ≤ ϕ(m) and rx ≡ α mod m is called the index (or discrete logarithm) of α to the base r modulo m and denoted by indrα.
In particular, rindra≡a modm.
Theorem 2.4. Letmbe a positive integer with the primitive rootr. Ifa, bare positive integers coprime to m and k is a positive integer, then
(i) indr1≡0 mod ϕ(m)
(ii) indr(ab)≡indra+ indrb mod ϕ(m) (iii) indrak ≡k·indra modϕ(m).
Theorem 2.5. If pis a prime number,α∈N, m is equal topα or2pα, (n, ϕ(m)) =d then the congruence
xn≡a mod m
has solution if and only if d divides indxa. In case of solvability the congruence equation has d solutions.
Fermat’s Little Theorem says: Let a be a nonzero integer, and let p -a be prime.
Then ap−1 ≡1 mod p.
Proposition 2.6. Let a, b, n ∈ Z with n 6= 0. The congruence ax ≡ b modn has solutions if and only if (a, n) | b. When the congruence has a solution x0 ∈ Z then the full solution set is {x(a,n)0+tn :t∈Z}.
It follows that the equation ax = b in Zn has (a, n) solutions. In particular, if (a, n) = 1, then the equation ax=b has unique solution in Zn.
2.7. Divisibility of binomial coefficients. (see [6]) In 1852, Kummer proved that if m and n are nonnegative integers and pis a prime number, then the largest power of p dividing m+nm
equals pc, where c is the number of carries when m and n are added in base p. Equivalently, the exponent of a prime p in nk
equals the number of nonnegative integers j such that the fractional part of k/pj is greater than the fractional part of n/pj. It can be deduced from the fact that nk
is divisible by n/gcd(n, k). Another fact: an integern ≥2 is prime if and only if all the intermediate binomial coefficients nk
, k = 1, . . . , n−1 are divisible by n.
2.8. p-adic numbers. LetQbe the field of rational numbers. Every rational number x6= 0 can be represented in the formx=pr nm, wherer, n∈Z,m is a positive integer, (p, n) = 1, (p, m) = 1 and p is a fixed prime number. The p-adic norm of x is given by
|x|p =
p−r for x6= 0 0 for x= 0.
This norm satisfies the so called strong triangle inequality
|x+y|p ≤max{|x|p,|y|p}, and this is a non-Archimedean norm.
The completion of Q with respect to p-adic norm defines the p-adic field which is denoted by Qp. Any p-adic number x 6= 0 can be uniquely represented in the canonical form
(2.1) x=pγ(x)(x0+x1p+x2p2+. . .),
where γ = γ(x) ∈ Z and xj are integers, 0 ≤ xj ≤ p−1, x0 >0, j = 0,1,2, ... (see more detail [7, 18, 16]). In this case |x|p =p−γ(x).
Theorem 2.9. [4], [7], [18] In order to the equation x2 =a, 06=a=pγ(a)(a0+a1p+ ...),0 ≤aj ≤p−1, a0 >0 has a solution x∈ Qp, it is necessary and sufficient that the following conditions are fulfilled:
i) γ(a) is even;
ii) a0 is a quadratic residue modulo p if p6= 2; a1 =a2 = 0 if p= 2.
In this paper we shall generalize this theorem.
3. Equation xq =a.
In this section we consider the equation xq = a in Qp, where p is a fixed prime number, q ∈N and a ∈Qp. Our goal is to find conditions under which the equation has a solution x∈Qp. The case q= 2 is well known (see Theorem 2.9), therefore we considerq >2.
We need the following
Lemma 3.1. The following are true (i)
(3.1)
∞
X
i=0
xipi
!q
=xq0 +
∞
X
k=1
qxq−10 xk+Nk(x0, x1, . . . , xk−1) pk, where x0 6= 0, 0≤xj ≤p−1, N1 = 0 and for k ≥2
(3.2)
Nk=Nk(x0, . . . , xk−1) = X
m0,m1,...,mk−1:
Pk−1
i=0mi=q, Pk−1 i=1imi=k
q!
m0!m1!. . . mk−1!xm00xm11. . . xmk−1k−1. (ii) Let q=p be a prime number, then p|Nk if and only if p-k.
Proof. (i) Formulas (3.1) and (3.2) easily can be obtained by using multinomial the- orem.
(ii) The following formula is known:
(3.3) p!
m0!m1!. . . mk−1! = m0
m0
m0+m1 m1
. . .
m0 +m1+· · ·+mk−1
mk−1
.
By the condition Pk−1
i=0 mi =p we get 0 ≤ mi ≤ p. Moreover if mi0 =p for some i0, then mi = 0 for alli6=i0. In this case from the conditionPk−1
i=1 imi =k we obtain k =i0p.
Assume p | k, i.e., k =pk0 for some k0, where 1 ≤ k0 < k −1. Putting mk0 = p and mi = 0 for i6=k0, we getNk(x0, . . . , xk−1) = xpk
0 +Sk, where
Sk= X
0≤m0,m1,...,mk−1<p:
Pk−1
i=0mi=p, Pk−1 i=1imi=k
p!
m0!m1!. . . mk−1!xm00xm11. . . xmk−1k−1.
Since xpk
0 is not a multiple of p (otherwise, if we assume that rp = pT for some r ∈ {2, . . . , p −1} and T ∈ N, then it follows that r divides T since p is a prime number. Hence, rp−1 = pT0, where T0 = T /r. By the same way we get rp−2 = pT00 and iterating the process finally we get 1 =pT˜, for some ˜T ∈N, which is not possible), butSkis divisible bypsince each coefficient ofSk contains (see formula (3.3)) a factor
p mi0
(with 0< mi0 < p) which is divisible byp thanks to the divisibility property of binomial coefficients mentioned in the previous section. Therefore p-Nk.
Assume p - k then by the arguments mentioned above we get mi < p, for all i= 0, . . . , k−1, therefore each term of Nk is divisible by pconsequently p|Nk. 3.1. THE CASE (q, p) = 1. In this subsection we are going to analyze under what conditions the equation xq = a has solution in Qp, when q and p are coprimes. In this case the following is true.
Theorem 3.2. Let q >2 and (q, p) = 1. The equation
(3.4) xq =a,
0 6=a =pγ(a)(a0+a1p+. . .), 0 ≤aj ≤ p−1, a0 6= 0, has a solution x∈ Qp if and only if
1) q divides γ(a);
2) a0 is a q residue mod p.
Proof. Necessity. Assume that equation (3.4) has a solution
x=pγ(x)(x0+x1p+. . .), 0≤xj ≤p−1, x0 6= 0, then
(3.5) pqγ(x)(x0+x1p+. . .)q =pγ(a)(a0+a1p+. . .), i.e., γ(a) = qγ(x) and a0 ≡xq0 mod p.
Sufficiency. Let a satisfies the conditions 1) and 2). We construct a solution x of equation (3.4) using the idea of reduction to canonical form of ap-adic number using a system of carries. Put
γ(x) = 1 qγ(a).
Then by the condition 2) and due to 1≤a0 ≤p−1 there exists x0 such that xq0 ≡a0 modp, 1≤x0 ≤p−1.
In other words, there existsM1(x0) such that xq0 =a0+M1(x0)p.
Using the notations of Lemma 3.1 due to the fact that qxq−10 is not a multiple of p, (see Proposition 2.6) there exists x1 such that
qxq−10 x1+N1(x0) +M1(x0)≡a1 mod p, 1≤x1 ≤p−1.
Therefore, there exists M2(x0, x1) such that qxq−10 x1 + N1(x0) + M1(x0) = a1 + M2(x0, x1)p.
Proceeding in this way, we find the existence of xn such that
qxq−10 xn+Nn(x0, . . . , xn−1) +Mn(x0, . . . , xn−1)≡an mod p, 1≤xn≤p−1.
and Mn+1(x0, . . . , xn) such that
(3.6) qxq−10 xn+Nn(x0, . . . , xn−1) +Mn(x0, . . . , xn−1) =an+Mn+1(x0, . . . , xn)p for any n∈N.
Now from Lemma 3.1 and equality (3.6) it follows that
∞
X
i=0
xipi
!q
=xq0+
∞
X
k=1
qxq−10 xk+Nk(x0, x1, . . . , xk−1) pk=
=a0+M1(x0)p+
∞
X
k=1
(ak−Mk(x0, . . . , xk−1) +Mk+1(x0, . . . , xk)p)pk =a0+
∞
X
k=1
akpk.
Hence, we found solution x=
∞
X
i=0
xipi of equation (3.4) in its canonical form.
Remark 3.3. The condition 2) of Theorem 3.2 is always satisfied if p = 2, conse- quently xq =a has a solution in Q2 for any odd q and any a∈Q2 with γ(a) divisible by q.
Corollary 3.4. Let q be a prime number such that q < p and η be a p-adic unit (i.e. |η|p = 1) which is not the q-th power of some p-adic number. Then piηj, i, j = 0,1, . . . , q−1 (i+j 6= 0) is not the q-th power of some p-adic number.
Proof. We shall check that the conditions of Theorem 3.2 fail under the hypothesis. It is easy to see thatγ(piηj) = ifor alli= 1, . . . , q−1 andj = 0, . . . , q−1, consequently q does not divide γ(piηj), i.e. the condition 1) is not satisfied. But for ηj, i = 0, j = 2, . . . , q−1 the condition 1) is satisfied, therefore we shall check the condition 2).
Consider the decomposition η=η0+η1p+. . ., then ηj =η0j+jηj−10 η1p+. . .. It is known (see Theorem 2.4) thatη0j ≡aq0 mod phas a solutionη0 if and only if inda0ηj0 is divisible by d= (p−1, q).
Sinceηis a unity which is not theq-th power of somep-adic number we haveη0 ≡aq0 mod phas not solution thus inda0η0 is not divisible by d= (p−1, q). This property implies that for a prime q with q < p the prime number p has a form p = qk+ 1.
Consequently, d= (p−1, q) = (qk, q) = q.
If inda0η0 = dl+r, then inda0η0j ≡ j(ql+r) mod (p−1), i.e., inda0η0j = j(ql+ r) +M(p−1) =q(jl+M k) +jr, but since 0< r < q−1, 2≤j ≤q−1 and q is a prime number, jr is not divisible by q. Thus inda0η0j is not divisible by d=q, hence
the condition 2) is not satisfied.
Corollary 3.5. Let q be a prime number such that q < p=qk+ 1, for some k ∈N andη be a unity which is not theq-th power of somep-adic number. Then any p-adic number x can be written in one of the following forms x=εijyijq, where εij ∈ {piηj : i, j = 0,1, . . . , q−1} and yij ∈Qp.
Proof. Let η = η0 +η1p+. . . and µ = µ0 +µ1p+. . . be unities which are not the q-th power of some p-adic numbers.
We shall show that there exists i ∈ {1, . . . , q −1} such that µ = ηiyq for some y∈Qp.
Note thatηi and η1i,i= 1, . . . , q−1 are not theq-th power of somep-adic numbers.
Consider µ = µ0 +µ1p+. . . and η1i = c0i+c1ip+. . ., then ηµi = µ0c0i + (µ1c0i + µ0c1i)p+. . .
It is easy to see that γ(µ/ηi) = 0, consequently the condition 1) of Theorem 3.2 is satisfied. Indeed, if µ0c0i =pM, then since p is a prime number, the last equality is possible if and only if µ0 and c0i divide M, consequently we get 1 =pM˜, which is impossible.
Now we shall check the condition 2) of Theorem 3.2.
The equation xq0 ≡µ0c0i mod phas a solution if and only if indx0µ0c0i is divisible by q = (q, p−1) (see Theorem 2.5). We have that c0i ≡ ci01 mod p, consequently indx0µ0c0i ≡ indx0µ0 +iindx0c01 mod (p−1). Assume indx0µ0 = s, indx0c01 = r, s, r = 1, . . . , q −1 and take i such that ir ≡ q−s mod (p−1) (the existence of i follows from the fact that (r, p− 1) = 1). It is clear that if s varies from 1 to q−1 then i also varies in {1, . . . , q−1}. Consequently we get indx0µ0c0i ≡ s+ir
mod (p−1)≡q mod (p−1). Hence the condition 2) in Theorem 3.2 is also satisfied, so there is a number isuch that µ=ηiyiq, for someyi ∈Qp.
For x ∈ Qp, let x = pγ(x)(x0 +x1p+. . .) and denote ν = x0 +x1p+. . . If ν satisfies conditions of Theorem 3.2, thenν =yq andx=pγ(x)yq. Ifν does not satisfy conditions of Theorem 3.2, then (as it was showed above) there exists i such that ν =ηiyq and in this case we get x=pγ(x)ηiyq. Taking γ(x) =qN +j completes the
proof.
Corollary 3.6. Let q be a prime number such that q < p 6=qk+ 1, for any k ∈ N. Then any p-adic number x can be written in one of the following forms x = εiyiq, where εi ∈ {pi :i= 0,1, . . . , q−1} and yi ∈Qp.
Proof. Using arguments in the proof of Corollary 3.5 we conclude that if p6=qk+ 1 then any unity is the q-th power of a p-adic number. Hence for x∈ Qp with γ(x) =
qN +i we get x=piyq, y∈Qp.
3.2. THE CASE q=p. Now we are going to analyze the solvability conditions for the equation xp =a inQp. In this case, the following is true.
Theorem 3.7. Let q = p. The equation xq = a, 0 6= a = pγ(a)(a0 +a1p+. . .), 0≤aj ≤p−1, a0 6= 0, has a solution x∈Qp if and only if
(i) p divides γ(a);
(ii) ap0 ≡a0+a1p modp2.
Proof. Necessity. Assume that equation (3.4) has a solution
x=pγ(x)(x0+x1p+. . .), 0≤xj ≤p−1, x0 6= 0, then using Lemma 3.1 we get
(3.7) a=ppγ(x)(x0+x1p+. . .)p =ppγ(x) xp0+
∞
X
k=1
(pxp−10 xk+Nk)pk
! .
Using part (ii) of Lemma 3.1, we get
RHS of (3.7) =ppγ(x)
xp0+
∞
X
k=1 p|k
xp−10 xkpk+1+
∞
X
k=1 p|k
Nkpk+
∞
X
k=1 p-k
(xp−10 xk+p−1Nk)pk+1
=
ppγ(x) xp0+
∞
X
k=1
xp−10 xpkppk+1+
∞
X
k=1
Npkppk+
p−1
X
i=1
∞
X
k=0
(xp−10 xpk+i+p−1Npk+i)ppk+i+1
!
=
ppγ(x) xp0+
p−2
X
i=1
(xp−10 xi+p−1Ni)pi+1+
∞
X
k=1
(xp−10 xpk−1+Npk+p−1Npk−1)ppk+
(3.8)
∞
X
k=1
xp−10 xpkppk+1+
p−2
X
i=1
∞
X
k=1
(xp−10 xpk+i+p−1Npk+i)ppk+i+1
! .
We have
Npk = X
m0,...,mpk−1:
Ppk−1
i=0 mi=p, Ppk−1 i=1 imi=pk
p!
m0!. . . mpk−1!xm00. . . xmpk−1pk−1 =
(3.9) =p(p−1)xp−20 x1xpk−1+ ˜Npk, where ˜Npk does not depend on xpk−1; moreover p-N˜pk.
Using (3.9), from (3.8) we get
RHS of (3.7) =ppγ(x) xp0 +
p−2
X
i=1
(xp−10 xi+p−1Ni)pi+1+
∞
X
k=1
(xp−10 xpk−1+ ˜Npk+p−1Npk−1)ppk+
∞
X
k=1
(xp−10 xpk−xp−20 x1xpk−1)ppk+1+
∞
X
k=1
(xp−10 xpk+1+xp−20 x1xpk−1+p−1Npk+1)ppk+2+
(3.10)
p−2
X
i=2
∞
X
k=1
(xp−10 xpk+i +p−1Npk+i)ppk+i+1
! .
Consequently, γ(a) = pγ(x), andxp0 ≡a0 +a1p mod p2, x0 =a0.
Sufficiency. Assume thata satisfies the conditions (i) and (ii). We shall construct a solutionxof the equation xp =ausing the similar process of reduction to canonical form as in the proof of Theorem 3.2. First put
γ(x) = 1 pγ(a).
Denote x0 =a0 and let M1 be such that xq0 =a0+a1p+M1p2.
Proceeding in this way, sincexp−10 is not a multiple ofpand taking into account that integer numbersNk(see Lemma 3.1) depend only onx0, x1, . . . , xk−1 and ˜Npk depends only on x0, x1, . . . , xpk−2 we find the existence of xn and introduce corresponding number Mn+1 consequently for each n≥1 such that the following congruences hold:
1)xp−10 xi +p−1Ni+Mi ≡ai+1 mod p,
therefore, there exists Mi+1 such that xp−10 xi +p−1Ni +Mi = ai+1 +Mi+1p for 0 ≤ xi ≤p−1, i= 1, . . . , p−2;
2)xp−10 xpk−1+ ˜Npk+p−1Npk−1+Mpk−1 ≡apk mod p,
therefore, there existsMpksuch thatxp−10 xpk−1+ ˜Npk+p−1Npk−1+Mpk−1 =apk+Mpkp for k = 1,2, . . .;
3)xp−10 xpk−xp−20 x1xpk−1+Mpk ≡apk+1 modp,
therefore, there existsMpk+1 such thatxp−10 xpk−xp−20 x1xpk−1+Mpk =apk+1+Mpk+1p for k = 1,2, . . .;
4)xp−10 xpk+1+xp−20 x1xpk−1 +p−1Npk+1+Mpk+1 ≡apk+2 modp,
therefore, there existsMpk+2 such thatxp−10 xpk+1+xp−20 x1xpk−1+p−1Npk+1+Mpk+1 = apk+2+Mpk+2p for k = 1,2, . . .;
5)xp−10 xpk+i+p−1Npk+i+Mpk+i ≡apk+i+1 modp,
therefore, there exists Mpk+i+1 such that xp−10 xpk+i +p−1Npk+i+Mpk+i =apk+i+1+ Mpk+i+1pfor i= 2, . . . , p−2, k = 1,2, . . .;
Now making the substitutions above into equality (3.10) we obtain RHS of (3.10) =ppγ(x) a0+a1p+M1p2+
p−2
X
i=1
(ai+1−Mi+Mi+1p)pi+1+
∞
X
k=1
(apk−Mpk−1+Mpkp)ppk+
∞
X
k=1
(apk+1−Mpk+Mpk+1p)ppk+1+
∞
X
k=1
(apk+2−Mpk+1+Mpk+2p)ppk+2+
p−2
X
i=2
∞
X
k=1
(apk+i+1−Mpk+i+Mpk+i+1p)ppk+i+1
!
=
ppγ(x)
∞
X
k=1
akpk+M1p2−
∞
X
k=1
(Mk−Mk+1p)pk+1
!
=pγ(a)
∞
X
k=1
akpk=a.
Hence, we found the solution x=
∞
X
k=0
xkpk of the equation xp =a.
Corollary 3.8. Let p be a prime number.
a) The numbers ε ∈ E1 = {1} ∪ {i+jp : ip is not equal i+jp modulo p2}, δ∈ E2 ={pj :j = 0, . . . , p−1} and productsεδ are not thep-th power of some p-adic numbers.
b) Any p-adic numberx can be represented in one of the following forms x=εδyp, for some ε∈ E1, δ ∈ E2 and y ∈Qp.
Proof. a) Follows from Theorem 3.7.
b) If x = x0 + x1p+ · · · 6= yp for any y ∈ Qp then by Theorem 3.7 we have ε =x0 +x1p∈ E1. We shall show that xε = x0+xx1p+x2p2+...
0+x1p =b0+b1p+b2p2+. . . is the p-th power of some y ∈ Qp, i.e., we check the conditions of Theorem 3.7: since γ(x/ε) = 0 the condition (i) is satisfied; we have x0 ≡ x0b0 modp and x0 +x1p ≡ x0b0 + (x0b1 +x1b0)p mod p2 which implies that b0 = 1 and b1 = 0, consequently bp0 ≡b0+b1 mod p2 i.e., the condition (ii) is satisfied.
Thus if x ∈Qp has the form x=yp, then ε =δ = 1. If x= x0+x1p+. . . is not thep-th power of somep-adic number withγ(x) =pN+j, then we takeε =x0+x1p,
δ=pj then x=εδyp for some y∈Qp.
Note that for a given i∈ {1, . . . , p−1} the congruence ip ≡i+jp mod p2 is not satisfied for some values of j. For example, ifp= 3, then forj = 1 the congruence is not true. Using computer we get the following
Values of j s.t. ip ≡i+jp modp2 has no solution i∈ {1, . . . , p−1}
p=3 1
p=5 2
p=7 1, 3, 5 p=11 1, 4, 5, 6, 9 p=13 2, 3, 4, 8, 9, 10 p=17 1, 5, 8, 11, 15 p=19 4, 7, 8, 9, 10, 11, 14
p=23 3, 4, 6, 9, 10, 12, 13, 16, 18, 19 p=29 3, 5, 10, 11, 12, 13, 15, 17, 18, 23, 25
p=31 1, 2, 5, 6, 8, 9, 11, 15, 19, 21, 22, 24, 25, 28, 29
p=37 1, 4, 5, 6, 7, 10, 13, 14, 16, 20, 22, 26, 29, 30, 31, 32, 35 p=41 2, 4, 6, 8, 10, 16, 24, 26, 30, 32, 34, 36, 38
Remark 3.9. Using this table and Corollary 3.8 we obtain
p= 3: Any x∈Q3 has form x=εδy3, where ε∈ {1,4,5}, δ ∈ {1,3,9};
p = 5: Any x ∈ Q5 has form x = εδy5, where ε ∈ {1,11,12,13,14}, δ ∈ {1,5,25,125,625};
p= 7: Anyx∈Q7 has formx=εδy7, whereε∈ {1,8,9,10,11,12,13,22,23,24,25, 26,27,28,36,37,38,39,40,41,42}, δ∈ {7i :i= 0,1, . . . ,6}.
Remark 3.10. In [17] using Krasners’s Lemma a set of irreducible polynomials that defines all cubic extension of the field Q3 are determined.
3.3. THE CASEq=mps. As it is presented above, the proofs of the sufficient part of the solvability criteria in Theorems 3.2 and 3.7 are given in a constructive method.
Thus we show not only the existence of a solution, but we also give the algorithm for the constructing of the solution in these cases. After cases 3.1 and 3.2 it remains the case q = mps with some m, s ∈ N, (m, p) = 1. Here we shall show that this case can be reduced to cases 3.1 and 3.2: we have to find the solvability condition for xmps = a. Denoting y = xps, we get ym = a, which is the equation considered in case 3.1. Assume for the last equation the solvability condition is satisfied and its solution is y = ˜y. Then we have to solve xps = ˜y; here we denote z =xps−1 and get zp = ˜y. The last equation is the equation considered in case 3.2. Suppose it has a solutionz = ˜z (i.e. the conditions of Theorem 3.7 are satisfied) then we getxps−1 = ˜z which again can be reduced to the case 3.2. Iterating the last argument after s−1 times we obtain xp = ˜a for some ˜a which is also an equation corresponding to case 3.2. Consequently, by this argument we establish solvability condition of equation xmps =a, which will be a system of solvability conditions for equations considered in cases 3.1 and 3.2.
Remark 3.11. Note that in [14] for the case 3.3. an explicit formula (condition) for the existence of solution is obtained.
Acknowledgements
The first author was supported by Ministerio de Ciencia e Innovaci´on (European FEDER support included), grant MTM2009-14464-C02-02. The second and third authors thank the Department of Applied Mathematics, E.U.I.T. Forestal, Univer- sity of Vigo, Pontevedra, Spain, for providing financial support of their visit to the Department.
We thank F.M.Mukhamedov and M.Saburov for useful discussions on previous version of our paper. Taking in to account their comments we improved the style of proofs of Theorems 3.2 and 3.7.
We thank both referees for their useful suggestions.
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J. M. Casas, Department of Applied Mathematics, E.U.I.T. Forestal, University of Vigo, 36005, Pontevedra, Spain.
E-mail address: [email protected]
B. A. Omirov, Institute of mathematics, Tashkent, Uzbekistan.
E-mail address: [email protected]
U. A. Rozikov, Institute of mathematics, Tashkent, Uzbekistan.
E-mail address: [email protected]
